The value of $n$ for which $704+\frac{1}{2}\left(704\right)+\frac{1}{4}\left(704\right)+\dots$up $n$ terms $=1984-\frac{1}{2}\left(1984\right)+\frac{1}{4}\left(1984\right)-....$ up to $n$ terms is

According to the given condition

(704) $\left\{1-\frac{1}{2^n}\right\}=\frac{\left(1984\right)\left(2\right)}{3}\left\{1-\frac{(-1{)}^n}{2^n}\right\}$(2)

$\Rightarrow 128=\frac{2112}{2^n}-1984\frac{(-1{)}^n}{2^n}$

If $n$ is odd, we get $2^n=32$ or $n=5$

If $n$ is even, we get $128=128/2^n\Rightarrow n=0.$